what is the diff. of sin6x+cos6x/sin3x+cos3x

Dear Mustafa,

Its diferentiation will be : -

( 6cos6x  - 6sin6x ) /(3cos3x - 3sin3x )

or, 2 ( cos6x - sin6x ) / ( cos3x - cos3x )

Thank you.

Dear Mustafa,

Its diferentiation will be : -

( 6cos6x  - 6sin6x ) /(3cos3x - 3sin3x )

or, 2 ( cos6x - sin6x ) / ( cos3x - cos3x )

Thank you.

Hi mustafa

y=(sin6x+cos6x)/(sin3x+cos3x)

dy/dx ={(sin3x+cos3x)(6cos6x-6sin6x)-(3cos3x-3sin3x)(sin6x+cos6x)}/(sin3x+cos3x)² 

^{Now simplyfy}

^{dy/dx =(3 cos9x -9sin3x)/} (sin3x+cos3x)²